Question

3 shows the probability density for an electron that has passed through an experimental apparatus. What is the probability that the electron will land in a 0.010-mm-wide strip at (a) x = 0.000 mm, (b) x = 0.500 mm, (c) x = 1.000 mm, and (d) x = 2.000 mm?

Short answer

Determine the probability of electron will land in a 0.010 mm wide string at a position of x=0.000 mm using the formula.

Step-by-step solution

The probability of any electron ends up in the strip at position x is

Prob (in $\delta x$at x)=p(x)$\delta$x.........(1)

Here p(x) is the probability density and $\delta$x is the small width

(a) Convert the units for the small width of the strip from mm to m.

From the figure EX39.13, the probability density of the electron at the position x=0.000 mm is as follows.

p(x)=0.50mm

From the fig . the probability density of the electron at the position x=0.000mm is follows.

p(x)= 0.50 mm^-1

= 0.50x10³ m^-1

Substitute for p(x) 1x10^-5m for $\delta x$in the equation(1) to solve for

prob(in 0.010mm at 0.000mm)

prob(in 0.010mm at 0.000mm)=(0.50x10³m^-1)(1x10^-3m)

=5.0x10^-3

Therefore the probability of electron will land in a 0.010 mm wide strong at a position of x=0.000mm is 5.0x10^-3

(b)Determine the probability of electron will land in a 0.010mm wide string at a position of x=0.500mm using the equation(1)

From the figure EX39.13, the probability density of the electron at the position x=0.000 mm is as follows.

p(x)=0.50mm

From the fig . the probability density of the electron at the position x=0.000mm is follows.

p(x)= 0.50 mm^-1

= 0.50x10³ m^-1

Substitute 0.50x10³ m^-1 for p(x) 1x10^-5m for $\delta x$in the equation(1) to solve for

prob(in 0.010mm at 0.000mm)

prob(in 0.010mm at 0.500mm)=(0.25x10³m^-1)(1x10^-3m)

=2.5x10^-3

Therefore the probability of electron will land in a 0.010 mm wide strong at a position of x=0.500mm is 2.5x10^-3

(c)Determine the probability of electron will land in a 0.010mm wide string at a position of x=1.000mm using the equation(1)

From the fig. the probability density of the electron at the position x=1.000mm is as follows

p(x)=0.50mm

From the fig . the probability density of the electron at the position x=0.000mm is follows.

p(x)= 0.0 mm^-1

= 0.0x10³ m^-1

Substitute 0.0m^-1for p(x) 1x10^-3 for $\delta x$in the equation (1) to solve for

prob(in 0.010mm at 1.000mm)

prob(in 0.010mm at 1.000mm)=(1.000x10³m^-1)(1x10^-3m)

= 1.000x10^-3

Therefore the probability of electron will land in a 0.010 mm wide strong at a position of x=1.000mm

(d)Determine the probability of electron will land in a 0.010mm wide string at a position of x=2.000mm using the equation(1)

From the fig. the probability density of the electron at the position x=1.000mm is as follows

p(x)=0.50mm

From the fig . the probability density of the electron at the position x=0.000mm is follows.

p(x)= 0.25 mm^-1

= 0.25x10³ m^-1

Substitute 0.25x10³ m^-1 for p(x) 1x10^-3 for 1x10^-5in the equation (1) to solve for

prob(in 0.010mm at 2.000mm)

prob(in 0.010mm at 2.000mm)=(0.25x10³m^-1)(1x10^-3m)

=2.5x10^-3

Therefore the probability of electron will land in a 0.010 mm wide strong at a position of x=2.000mm is 2.5x10^-3